Which of the following coordination compounds is not likely to show...

Optical isomerism is generally shown by octahedral complexes that lack elements of symmetry (plane of symmetry, center of symmetry).

Let us analyze each complex:

Option (1): $[NiCl_2(en)_2]$ can exist in cis and trans isomeric forms. The cis form lacks a plane of symmetry and is optically active. The trans form has a plane of symmetry and is optically inactive. Therefore, the complex as a whole can show optical isomerism.

Option (2): $[Co(en)_3]^{3+}$ is an octahedral complex with three bidentate ligands. It lacks any plane or center of symmetry. Thus, it exists as a pair of non-superimposable mirror images (enantiomers) and shows optical isomerism.

Option (3): $[Cu(NH_3)_4]^{2+}$ is a typical example of a square planar complex. Square planar complexes contain a molecular plane of symmetry (the plane containing the metal ion and the four ligands). Because of this plane of symmetry, they do not show optical isomerism.

Option (4): $[PtCl_2(en)_2]^{2+}$ is an octahedral complex similar to the first option. Its cis-isomer is optically active, while the trans-isomer is optically inactive. Therefore, it can show optical isomerism.

Thus, the square planar complex $[Cu(NH_3)_4]^{2+}$ is not likely to show optical isomerism.